For a compound event $E_1E_2$,
Pr($E_1E_2$) = Pr($E_1$)Pr($E_2$|$E_1$)
such that if $E_1$ and $E_2$ are independent events, we can say that:
Pr($E_1E_2$) = Pr($E_1$)Pr($E_2$)

Is there an analogous formula that can be derived for the conditional probability: Pr($E_1E_2$|$E_3$)?

I would like to be able to reduce Pr($E_1E_2$|$E_3$) to terms that do not involve all three of the events $E_1$, $E_2$, $E_3$, given the assumption that only $E_1$ and $E_3$ are independent.

If this cannot be done, then what assumptions would be necessary in order to be able to do it?


Here is one way to write it given the assumption: $$ P(E_1\cap E_2\mid E_3)=P(E_2\mid E_1\cap E_3)P(E_1\mid E_3)=P(E_2\mid E_1\cap E_3)P(E_1). $$

  • $\begingroup$ Is there no way to re-write it as a product of terms, each of which only involves one or two out of the three events? $\endgroup$ – user1247 Jan 28 '13 at 10:40
  • $\begingroup$ I don't think there is, but I am not sure. You should probably edit your question and state you're looking for such an expression. $\endgroup$ – Stefan Hansen Jan 28 '13 at 10:45
  • $\begingroup$ OK I edited my question. I guess this boils down to something like: if A and B are independent, can I write P(A|BC)=P(A|C)? If not, what are the minimal assumptions that are necessary in order for that to be true? $\endgroup$ – user1247 Jan 28 '13 at 10:55

Hint: If you write out $$Pr(E2|E1)=\frac{Pr(E_2\cap E_1)}{Pr(E_1)}$$ then you can investigate more complicated formulas involving more events. Don't remember a formula but understand them.

  • $\begingroup$ The problem is that I can't figure out how using that formula would enable me to show, for example, that P(A|BC)=P(A|C), given A and B being independent. $\endgroup$ – user1247 Jan 28 '13 at 12:32

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